Does Cantor's paradox require the axiom of choice?

Question

Wikipedia's article about Cantor's paradox claims that it requires that the cardinals are linearly ordered. But can't we show that the cardinality of the universal set is the greatest cardinality using the inclusion maps of the other sets? Can't we define the greatest cardinality even if the cardinalities are not well-ordered? The definition of a greatest element works in partial orders, not just total orders. Also, can Cantor's paradox prove that there is no maximal cardinality even if the cardinalities are not linearly ordered?

Answer 1

The Wikipedia article, as written currently on 26 April, 2023 (14:51 UTC) is absolutely wrong about this.

You can talk about a greatest element of a partial order which is not well-ordered, for example $\mathcal P(X)$ has a greatest element under $\subseteq$, $X$ itself, even though when $X$ has at least two elements, this partial order is not a total order.

Moreover, given any set of cardinals we can find a cardinal which is strictly larger than all of them. Which means that at the very least the partial order of cardinals, even in the absence of $\sf AC$, is directed (indeed, very directed).¹

Indeed, if one thinks about it closely, $|V_\alpha|$ ranging over the ordinals, is a linearly ordered and cofinal class of cardinals. So it is enough to argue that this chain does not have an upper bound. But this, of course, requires a slightly more subtle understanding of the universe of sets.

This, on its own, will already give you that there is no maximal cardinality. As any cardinality must be below some $|V_\alpha|$, a maximal cardinality would have to be a $|V_\alpha|$. But then by Cantor's theorem/paradox, $V_{\alpha+1}$ cannot exist.

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Footnotes.

1. If $A$ is a set of sets, then $\bigcup A$ is at least as large as each of those, and therefore $\mathcal P(\bigcup A)$ has a larger cardinality. If, however, one thinks of cardinals as Scott cardinals (and/or initial ordinals for the well-orderable sets), this is also not a problem, since $\bigcup A$ is now a set of sets which has the relevant cardinalities (adjusting for initial ordinals, but one can see that this is not really a problem), and therefore $\mathcal P(\bigcup\bigcup A)$ works. If one assumes $\sf ZF$, then one can simply take a large enough $V_\alpha$ to function as a set with the larger cardinality.